An element $p$ of a commutative ring $R$ is called "prime" if, for any $a,b\in R$, whenever $ab$ is a multiple of $p$, either $a$ or $b$ is a multiple of $p$.

Is there a word for the "prime-like" property that, whenever $ab$ is a multiple of $p^2$, either $a$ or $b$ is divisible by $p$? Or another, more usual concept in ring theory that this is connected to?

I ask because the "prime-likeness" of $2$ in $R$ seems to control whether the quadratic formula can be made to work for monic polynomials over $R$ (as long as $2$ is also not a zero-divisor). This is because, if the discriminant of $x^2 + bx + c$ is a square $b^2 - 4c = d^2$, then $(-b+d)(-b-d) = 4c$, so at least one (and hence both) of $(-b+d)$ and $(-b-d)$ are multiples of $2$ in $R$. Their halves are the two roots of $x^2 + bx + c$.

For example, $2$ is "prime-like" in $\mathbb{Z}[\sqrt{2}]$, which is easy to verify elementarily. Hence a monic quadratic over $\mathbb{Z}[\sqrt{2}]$ factors iff its discriminant is a square. But $2$ is not "prime-like" in $\mathbb{Z}[\sqrt{5}]$, since $(\sqrt{5}-1)(\sqrt{5}+1) = 4$. And indeed, the discriminant of $x^2 -x-1$ is a square in $\mathbb{Z}[\sqrt{5}]$, but the polynomial doesn't factor there.

1 Answer
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I assume (based on your example) that you're primarily interested in the case where $R$ is the ring of integers in an algebraic extension of $\mathbb Q$. Then your property of being "prime-like" is equivalent to the property of generating a primary ideal.

So assume $R$ is a Dedekind domain (in particular ideals factor uniquely into products of prime ideals), and let $p\in R$ be an arbitrary element. Then

$p$ is "prime-like" if and only if $pR=P^k$ for some prime ideal $P$ of $R$ and some $k\in \mathbb N$

i.e. "prime-like" in Dedekind domains is the same as "primary". An element $x$ in the fraction field of $R$ lies in $R$ iff the valuations $\nu_Q(x)$ are non-negative for all prime ideals $Q$ of $R$. So let $a,b\in R$ and assume $p^2\mid ab$. Then $\nu_Q(a/p)=\nu_Q(a)\geq 0$ and $\nu_Q(b/p)=\nu_Q(b)\geq 0$ for all primes $Q\neq P$. So to see that either $a/p$ or $b/p$ lies in $R$, one just has to check that one of them has positive $P$-valuation. But

so either $\nu_P(a/p)\geq 0$ or $\nu_P(b/p)\geq 0$. On the other hand, if the ideal $pR$ isn't primary then $p$ is not prime-like (the construction in Julian's comment can be generalized).

Of course I'm not sure what exactly you're looking for, but at least this clears up what is going on in your last example: while $2$ isn't "prime-like" in $\mathbb Z[\sqrt{5}]$, it is prime like in its integral closure $\mathbb Z[\frac{1+\sqrt{5}}{2}]$ (which is however unspectacular because it remains a prime in that ring).

Certainly all primes would be "primelike" in any ring. Let $p$ be prime. If $p^2 \mid ab$, then $ab \in (p^2) \subseteq (p)$ which implies $a \in (p)$ or $b \in (p)$. Don't think you need valuations for that. I am curious though where it stands in relation to irreducible or pseudo-prime or something like this.
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CPMAug 18 '12 at 23:33

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@CPM: The point is that $p$ is a prime in $\mathbb Z$, but not necessarily in a ring of algebraic integers (namely if it's ramified at the extension). A main point of the question seems to be to determine whether $2$ is prime-like in certain extensions of $\mathbb Z$, so I think it's definitely reasonable to focus attention on primes in $\mathbb Z$.
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Florian EiseleAug 18 '12 at 23:43

@CPM: To be even shorter: It is most definitely false that any prime $p\in\mathbb Z$ is prime-like in any ring.
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Florian EiseleAug 18 '12 at 23:52

Ah, I thought he was wondering in general about where prime-like stood in relation to prime in any old ring. My mistake.
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CPMAug 18 '12 at 23:54

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In general, there are many ways to extend the valuation $v_p$ to the ring of integers of a number fields (the extension correspond to primes lying over $p$). I believe your statement that a prime $p\in\mathbb{Z}$ is prime-like in the ring of algebraic integers in a number field is false. Consider, for example, $p=5$, $R=\mathbb{Z}[i]$, $a=(2+i)^2$, $b=(2-i)^2$. We have $ab=25$, so $5^2|ab$, but $5\nmid a$, $5\nmid b$.
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Julian RosenAug 19 '12 at 0:22